Question

Two investments totaling $56,000 produce an annual income of $1400. One investment yields 4% per year, while the other yields 2% per year. How much is invested at each rate?

Answer 1

Solution:

Let x represent one of the investments

Since the total of the two investments is $56,000, the other investment will be

$=56000-x$

If x is invested at 4% and the other is invested at 2%, then, it can be expressed as

$\begin{gathered} 4\%\text{ of x}=\frac{4}{100}\times x=0.04x \\ 2\%\text{ of \lparen56000-x\rparen}=\frac{2}{100}\times(56000-x)=0.02(56000-x)=1120-0.02x \end{gathered}$

And, their income is $1400, then, it can be expressed as

$0.04x+(1120-0.02x)=1400$

Solve for x

$\begin{gathered} 0.04x+(1120-0.02x)=1400 \\ 0.04x+1120-0.02x=1400 \\ Collect\text{ like terms} \\ 0.04x-0.02x=1400-1120 \\ 0.02x=280 \\ Divide\text{ both sides by 0.02} \\ \frac{0.02x}{0.02}=\frac{280}{0.02} \\ x=\text{\ 14000} \end{gathered}$

The other investment will be

$=56000-14000=\text{\ 42000}$

Hence,

At a 4% rate, the investment is $14000 and at a 2% rate, the investment is $42000